Question

An aeroplane is moving in a horizontal circle with a uniform speed of 141 km/hr.
Change in its velocity from its initial position will be

a)100km/hr at an angle of 90°
b)141km/hr at an angle of 135°
c)200km/hr at an angle of 135°
d)Zero​

Answer 1

An aeroplane is moving in a horizontal circle with a uniform speed of 141 km/hr.

Change in its velocity in one fourth revolution, from its initial direction will be

a) 100km/hr at an angle of 90°

b) 141km/hr at an angle of 135°

c) 200km/hr at an angle of 135°

d) Zero

solution : let the initial position of particle is at a point of positive direction of X - axis and it moves along y - axis.

so, initial velocity of aeroplane, u = 141 j

now one fourth revolution,

see diagram, final position of particle is at point P and velocity is along negative x - axis.

i.e., v = -141i

so, change in velocity = v - u

= -141i - 141j

magnitude of change in its velocity, ∆v = 141√(1² + 1²) ≈ 200 km/hr.

angle made with its initial velocity,

cosθ = (-141i - 141j)(141j)/|-141i - 141j||141j|

⇒cosθ = -141²/141²√2 = -1/√2

⇒θ = 135°

Therefore change in velocity from its initial position will be 200 km/hr at an angle 135°. hence option (c) is correct choice.